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NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables

Edited By Ramraj Saini | Updated on Sep 26, 2023 07:50 PM IST

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables

Linear Equations in Two Variables Class 9 Questions And Answers are provided here. These NCERT solutions are prepared by expert team at careers360 keeping in mind latest CBSE syllabus 2023. In this particular chapter, you will study linear equations in two variables of the type ax + by + c = 0 where a, b and c are the real numbers, and a and b both are not zero. Class 9 linear equations in two variables NCERT solutions are there to make your task easy while preparing for the exams.

This Story also Contains
  1. NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables
  2. Linear Equations in Two Variables Class 9 Questions And Answers PDF Free Download
  3. Linear Equations in Two Variables Class 9 Solutions - Important Formulae
  4. Linear Equations in Two Variables Class 9 NCERT Solutions (Intext Questions and Exercise)
  5. Linear Equations In Two Variables Class 9 NCERT Solutions - Summary
  6. NCERT Solutions for Class 9 - Subject Wise
  7. NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables - Key Features
  8. NCERT Books and NCERT Syllabus
NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables
NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables

This class 9 linear equations in two variables always comes with a good number of questions in competitive exams like the Indian National Olympiad (INO), National Talent Search Examination (NTSE). Linear equations in two variables class 9 NCERT solutions are designed in such a manner that a student can get maximum marks assigned to that particular question. NCERT solutions for class 9 are also available chapter wise which can be downloaded by clicking on the given link.

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Linear Equations in Two Variables Class 9 Questions And Answers PDF Free Download

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Linear Equations in Two Variables Class 9 Solutions - Important Formulae

  • (a + b)2 = a2 + 2ab + b2

  • (a - b)2 = a2 - 2ab + b2

  • (a + b)(a - b) = a2 - b2

  • (x + a)(x + b) = x2 + (a + b)x + ab

  • (x + a)(x - b) = x2 + (a - b)x - ab

  • (x - a)(x + b) = x2 + (b - a)x - ab

  • (x - a)(x - b) = x2 - (a + b)x + ab

  • (a + b)3 = a3 + b3 + 3ab(a + b)

  • (a - b)3 = a3 - b3 - 3ab(a - b)

  • (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz

  • (x + y - z)2 = x2 + y2 + z2 + 2xy - 2yz - 2xz

  • (x - y + z)2 = x2 + y2 + z2 - 2xy - 2yz + 2xz

  • (x - y - z)2 = x2 + y2 + z2 - 2xy + 2yz - 2xz

  • x3 + y3 + z3 - 3xyz = (x + y + z)(x2 + y2 + z2 - xy - yz - xz)

  • x2 + y2 = 1/2 [(x + y)2 + (x - y)2]

  • (x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc

  • x3 + y3 = (x + y)(x2 - xy + y2)

  • x3 - y3 = (x - y)(x2 + xy + y2)

  • x2 + y2 + z2 - xy - yz - zx = 1/2 [(x - y)2 + (y - z)2 + (z - x)2]

Free download NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables for CBSE Exam.

Linear Equations in Two Variables Class 9 NCERT Solutions (Intext Questions and Exercise)

Linear equations in two variables class 9 solutions Exercise: 4.1

Q1 The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement. (Take the cost of a notebook to be Rs x and that of a pen to be Rs y ).

Answer:

Let the cost of a notebook be Rs x and that of a pen be Rs y .

According to the given condition: The cost of a notebook is twice the cost of a pen.

Thus, x=2y

\Rightarrow x-2y=0

Q2 (i) Express the following linear equations in the form ax + by + c = 0 and indicate the values of a , b and c in each case: 2x + 3y = 9.3\bar{5}

Answer:

Given : 2x + 3y = 9.3\bar{5}

\Rightarrow 2x + 3y - 9.3\bar{5}=0

Here , a=2, b=3 and c = - 9.3\bar{5}

Q2 (ii) Express the following linear equations in the form ax + by + c = 0 and indicate the values of a , b and c in each case: x - \frac{y}{5} - 10 = 0

Answer:

Given:

x - \frac{y}{5} - 10 = 0

\Rightarrow x - \frac{y}{5} - 10 = 0

Here ,

a=1,

b=\frac{-1}{5}

c = -10

Q2 (iii) Express the following linear equations in the form ax + by + c = 0 and indicate the values of a , b and c in each case: -2x + 3y = 6

Answer:

Given :

-2x + 3y = 6

\Rightarrow -2x + 3y - 6=0

Here , a= -2, b=3 and c = -6

Q2 (iv) Express the following linear equations in the form ax + by + c = 0 and indicate the values of a , b and c in each case: x =3y

Answer:

Given : x =3y

\Rightarrow x -3y=0

Here , a= 1, b= -3 and c =0

Q2 (v) Express the following linear equations in the form ax + by + c = 0 and indicate the values of a , b and c in each case: 2x = - 5y

Answer:

Given : 2x = - 5y

\Rightarrow 2x + 5y=0

Here , a=2, b= 5 and c =0

Q2 (vi) Express the following linear equations in the form ax + by + c = 0 and indicate the values of a , b and c in each case: 3x + 2 = 0

Answer:

Given : 3x + 2 = 0

\Rightarrow 3x + 2 = 0

Here , a= 3, b=0 and c =2

Q2 (vii) Express the following linear equations in the form ax + by + c = 0 and indicate the values of a , b and c in each case: y -2 = 0

Answer:

Given : y -2 = 0

\Rightarrow 0.x+y -2 = 0

Here , a=0, b= 1 and c = -2

Q2 (viii) Express the following linear equations in the form ax + by + c = 0 and indicate the values of a , b and c in each case: 5 = 2x

Answer:

Given : 5 = 2x

\Rightarrow 2x+0.y-5=0

Here , a=2, b= 0 and c = -5

Class 9 maths chapter 4 question answer Exercise: 4.2

Q1 Which one of the following options is true, and why? y = 3x + 5 has

(i) a unique solution,

(ii) only two solutions,

(iii) infinitely many solutions

Answer:

Given : y = 3x + 5

This equation is of a line and a line has infinite points on it and each point is a solution Thus, (iii) infinitely many solutions is the correct option.

Q2 Write four solutions for each of the following equations:

(i) 2x + y = 7 (ii) \pi x +y = 9 (iii) x = 4y

Answer:

(i) Given : 2x + y = 7

Putting x=0, we have , y=7-2\times 0=7 means (0,7) is a solution.

Putting x=1, we have , y=7-2\times 1=5 means (1,5) is a solution.

Putting x=2, we have , y=7-2\times 2=3 means (2,3) is a solution.

Putting x=3, we have , y=7-2\times 3=1 means (3,1) is a solution.

The four solutions are : (0,7),(1,5),(2,3),(3,1) .

(ii) Given : \pi x +y = 9

Putting x=0, we have , y=9-\pi \times 0=9 means (0,9) is a solution.

Putting x=1, we have , y=9-\pi \times 1=9-\pi means (1,9-\pi ) is a solution.

Putting x=2, we have , y=9-\pi \times 2=9-2\pi means (2,9-2\pi ) is a solution.

Putting x=3, we have , y=9-\pi \times 3=9-3\pi means (3,9-3\pi ) is a solution.

The four solutions are : (0,9),(1,9-\pi ),(2,9-2\pi ),(3,9-3\pi ) .


(iii) Given : x = 4y

Putting x=0, we have , y=\frac{0}{4}=0 means (0,0) is a solution.

Putting x=1, we have , y=\frac{1}{4} means (1,\frac{1}{4}) is a solution.

Putting x=2, we have , y=\frac{2}{4}=\frac{1}{2} means (2,\frac{1}{2}) is a solution.

Putting x=3, we have , y=\frac{3}{4} means (3,\frac{3}{4}) is a solution.

The four solutions are : (0,0) , (1,\frac{1}{4}) , (2,\frac{1}{2}) and (3,\frac{3}{4}) .

Q3 (i) Check which of the following are solutions of the equation x - 2y = 4 and which are not: ((0,2)

Answer:

(i) Given : x - 2y = 4

Putting (0,2) ,

we have , x - 2y = 0-2(2)=-4\neq 4

Therefore, (0,2) is not a solution of x - 2y = 4 .

Q3 (ii) Check which of the following are solutions of the equation x - 2y = 4 and which are not: (2,0)

Answer:

Given : x - 2y = 4

Putting (2,0),

we have , x - 2y = 2-2(0)=2\neq 4

Therefore, (2,0) is not a solution of x - 2y = 4 .

Q3 (iii) Check which of the following are solutions of the equation x - 2y = 4 and which are not: (4,0)

Answer:

Given : x - 2y = 4

Putting (4,0),

we have , x - 2y = 4-2(0)=4=4

Therefore, (4,0) is a solution of x - 2y = 4 .

Q3 (iv) Check which of the following are solutions of the equation x - 2y = 4 and which are not: (\sqrt2 , 4\sqrt2)

Answer:

Given : x - 2y = 4

Putting (\sqrt2 , 4\sqrt2) ,

we have , x - 2y =\sqrt{2}-2(4\sqrt{2})=\sqrt{2}-8\sqrt{2}=-7\sqrt{2}\neq 4

Therefore, (\sqrt2 , 4\sqrt2) is not a solution of x - 2y = 4 .

Q3 (v) Check which of the following are solutions of the equation x - 2y = 4 and which are not: (1,1)

Answer:

Given : x - 2y = 4

Putting (1,1) ,

we have , x - 2y = 1-2(1)=-1\neq 4

Therefore, (1,1) is not a solution of x - 2y = 4 .

Q4 Find the value of k , if x = 2 , y = 1 is a solution of the equation 2x + 3y = k .

Answer:

Given : 2x + 3y = k

Putting (2,1),

we have , k=2x + 3y = 2(2)+3(1)=4+3=7

Therefore, k=7 for 2x + 3y = k putting x=2 and y=1.

Class 9 maths chapter 4 question answer Exercise: 4.3

Q1 (i) Draw the graph of each of the following linear equations in two variables: x + y = 4

Answer:

Given : x + y = 4

1640075509869

Putting x=0,we have y=4-0=4

Putting x=1,we have y=4-1=3

Thus, (0,4) and (1,3) are solutions of given equation.

Q1 (ii) Draw the graph of each of the following linear equations in two variables: x - y = 2

Answer:

Given : x - y = 2

1640075537340

Putting x=0,we have y=0-2=-2

Putting x=1,we have y=1-2=-1

Thus, (0,-2) and (1,-1) are solutions of given equation.

Q1 (iii) Draw the graph of each of the following linear equations in two variables: y = 3x

Answer:

Given : y = 3x

1640075560924

Putting x=0,we have y=3(0)=0

Putting x=1,we have y=3(1)=3

Thus, (0,0) and (1,3) are solutions of given equation.

Q1 (iv) Draw the graph of each of the following linear equations in two variables: 3 = 2x + y

Answer:

Given : 3 = 2x + y

1640075579373

Putting x=0,we have y=3-2.(0)=3

Putting x=1,we have y=3-2.(1)=1

Thus, (0,3) and (1,1) are solutions of given equation.

Q2 Give the equations of two lines passing through (2, 14). How many more such lines are there, and why?

Answer:

The equations of two lines passing through (2, 14) are given by : x+y=16\, \, and\, \, x-y=-12

There are infinite lines passing through (2, 14) because infinite lines pass through a point.

Q3 If the point (3, 4) lies on the graph of the equation 3y = ax+ 7 , find the value of a .

Answer:

Given : the point (3, 4) lies on the graph of the equation 3y = ax+ 7

Put x=3 and y=4

3y = ax+ 7

\Rightarrow 3(4) = a(3)+ 7

\Rightarrow 12 = 3a+ 7

\Rightarrow 12 -7 = 3a

\Rightarrow 5 = 3a

\Rightarrow a=\frac{5}{3}

Q4 The taxi fare in a city is as follows: For the first kilometre, the fare is Rs. 8 and for the subsequent distance it is Rs. 5 per km. Taking the distance covered as x km and total fare as Rs. y, write a linear equation for this information, and draw its graph.

Answer:

Given: The distance covered as x km and total fare is Rs. y.

Total fare =the fare for first km + the fare of the remaining distance

\therefore y=8+5\times (x-1)

\Rightarrow y=8+5x-5

\Rightarrow y=3+5x

For graph,

Putting x=0, we have y=3+5(0)=3

Putting x=1, we have y=3+5(1)=8

Putting x=2, we have y=3+5(2)=13

Hence,(0,3),(1,8) and (2,13) are solutions of equation.

The graph is as shown :

1640075605267

Q5 (A) From the choices given below, choose the equation whose graph is given in Fig. 4.6.

(i) y = x

(ii) x + y = 0

(iii) y = 2x

(iv) 2 + 3 y = 7x

1640075631362

Answer:

For the given figure :

Points on line are (-1,1) , (0,0 ) and (1,-1)

x + y = 0 satisfies all the above points.

Thus, x + y = 0 is the correct equation of the line.

Q5 (B) From the choices given below, choose the equation whose graph is given in Fig. 4.7

(i) y = x + 2

(ii) y = x - 2

(iii) y = -x + 2

(iv) x + 2y = 6


1640075653339

Answer:

For the given figure :

Points on line are (-1,3) , (0,2 ) and (2,0)

y = -x + 2 satisfies all the above points.

Thus, y = -x + 2 is the correct equation of the line.

Q6 (i) If the work done by a body on application of a constant force is directly proportional to the distance travelled by the body, express this in the form of an equation in two variables and draw the graph of the same by taking the constant force as 5 units. Also read from the graph the work done when the distance travelled by the body is: 2 units

Answer:

Let work done be y and distance be x.

Given : Constant force = 5 units

Work done by a body on the application of a constant force is directly proportional to the distance travelled by the body.

i.e. y\, \, \alpha \, \, x

\Rightarrow y=kx

k=Constant force = 5

Then,

\Rightarrow y=5x

For graph,

Put x=0,we have y=5(0)=0

Put x=1,we have y=5(1)=5

Put x=2,we have y=5(2)=10

Points are (0,0) , (0,5) and (2,10)

If the distance travelled is 2 units then the work done is 10 units.

1640075688295

Q6 (ii) If the work done by a body on application of a constant force is directly proportional to the distance travelled by the body, express this in the form of an equation in two variables and draw the graph of the same by taking the constant force as 5 units. Also read from the graph the work done when the distance travelled by the body is : 0 unit

Answer:

Let work done be y and distance be x.

Given : Constant force = 5 units

Work done by a body on the application of a constant force is directly proportional to the distance travelled by the body.

i.e. y\, \, \alpha \, \, x

\Rightarrow y=kx

k=Constant force = 5

Then, \Rightarrow y=5x

For graph,

Put x=0,we have y=5(0)=0

Put x=1,we have y=5(1)=5

Put x=2,we have y=5(2)=10

Points are (0,0) , (0,5) and (2,10)

If the distance travelled is 0 units then work done is 0 units.

Q7 Yamini and Fatima, two students of Class IX of a school, together contributed Rs. 100 towards the Prime Minister’s Relief Fund to help the earthquake victims. Write a linear equation which satisfies this data. (You may take their contributions as Rs. x and Rs. y.) Draw the graph of the same.

Answer:

Let the contribution of Yamini be x.

contribution of Yamini be y.

According to question,

x+y=100

For x=0 , we have y=100-0=100

For x=10 , we have y=100-10=90

For x=20 , we have y=100-20=80

Hence, (0,100) , (10,90) and (20,80)

1640075759528

Q8 (i) In countries like the USA and Canada, the temperature is measured in Fahrenheit, whereas in countries like India, it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius: F = \left(\frac{9}{5} \right )C + 32

Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit for the y-axis.

Answer:

Let celsius be on x-axis and Fahrenheit be on y-axis.

F = \left(\frac{9}{5} \right )C + 32

For graph,

Putting x=0, we get y= \left(\frac{9}{5} \right )(0)+ 32=32

Putting x=5, we get y= \left(\frac{9}{5} \right )(5)+ 32=41

Putting x=10, we get y= \left(\frac{9}{5} \right )(10)+ 32=50

Hence, points are (0,32) , (5,41) and (10,50).

1649229364736

Q8 (ii) In countries like USA and Canada, temperature is measured in Fahrenheit, whereas in countries like India, it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius: F = \left(\frac{9}{5} \right )C + 32

If the temperature is 30°C, what is the temperature in Fahrenheit?

Answer:

F = \left(\frac{9}{5} \right )C + 32

Put c=30,

F = \left(\frac{9}{5} \right )30 + 32=54+32=86

Thus, the temperature = 30°C, then the temperature is 86 in Fahrenheit.

Q8 (iii) In countries like USA and Canada, temperature is measured in Fahrenheit, whereas in countries like India, it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius: F = \left(\frac{9}{5} \right )C + 32

If the temperature is 95°F, what is the temperature in Celsius?

Answer:

F = \left(\frac{9}{5} \right )C + 32

Put F=95,

95 = \left(\frac{9}{5} \right )C + 32

\Rightarrow \left(\frac{9}{5} \right )C=95 -32

\Rightarrow \left(\frac{9}{5} \right )C=63

\Rightarrow C=35

If the temperature is 95°F, then 35 is the temperature in Celsius.

Q8 (iv) In countries like USA and Canada, temperature is measured in Fahrenheit, whereas in countries like India, it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius: F = \left(\frac{9}{5} \right )C + 32

If the temperature is 0°C, what is the temperature in Fahrenheit and if the temperature is 0°F, what is the temperature in Celsius?

Answer:

If the temperature is 0°C,

F = \left(\frac{9}{5} \right )C + 32

F = \left(\frac{9}{5} \right )(0) + 32

\Rightarrow F = 32

if the temperature is 0°F,

F = \left(\frac{9}{5} \right )C + 32

0= \left(\frac{9}{5} \right )C + 32

\Rightarrow \left(\frac{9}{5} \right )C=- 32

\Rightarrow C=- 17.8

Thus, if the temperature is 0°C , then the temperature in Fahrenheit is 32 and if the temperature is 0°F, then the temperature in celsius is -17.8.

Q8 (v) In countries like USA and Canada, temperature is measured in Fahrenheit, whereas in countries like India, it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius: F = \left(\frac{9}{5} \right )C + 32

(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.

Answer:

F = \left(\frac{9}{5} \right )C + 32

Let temperature be x in both Fahrenheit and celsius.

x= \left(\frac{9}{5} \right )x + 32

\Rightarrow 5x= 9x + 160

\Rightarrow 5x-9x = 160

\Rightarrow -4x = 160

\Rightarrow x = -40 \degree

Thus, 40 is the temperature which is numerically the same in both Fahrenheit and celsius.

Class 9 maths chapter 4 NCERT solutions Exercise: 4.4

Q1 (i) Give the geometric representations of y = 3 as an equation

in one variable

Answer:

Equation y = 3 can be represented in one variable on the number line.

1640075843993

Q1 (ii) Give the geometric representations of y = 3 as an equation: in two variables

Answer:

For 2 variable representations of y = 3 .

Equation : 0.x+y=3

For graph,

x=0,we have y=3

x=1,we have y=3

x=2,we have y=3

Hence, (0,3) ,(1,3) and (2,3) are solutions of equation.

1640075877910

Q2 (i) Give the geometric representations of 2x + 9 = 0 as an equation: in one variable

Answer:

Equation 2x + 9 = 0 can be represented in one variable on the number line.

\Rightarrow x=\frac{-9}{2}

The yellow mark represents x= - 4.5.

1640075916094

Q2 (ii) Give the geometric representations of 2x + 9 = 0 as an equation: in two variables

Answer:

For 2 variable representation of 2x + 9 = 0 .

Equation : 2x +0.y =-9

For graph,

y=0,we have x=\frac{-9}{2}

y=1,we have x=\frac{-9}{2}

y=2,we have x=\frac{-9}{2}

Hence, (\frac{-9}{2},0) , (\frac{-9}{2},1) and (\frac{-9}{2},2) are solutions of equation.

1649229419769

Linear Equations In Two Variables Class 9 NCERT Solutions - Summary

The chapter 4 maths class 9 Textbook is called "Linear Equations in Two Variables," which is included in Unit 2 Algebra. In the board exams, Algebra unit comprises a total of 20 marks, consisting of 1 multiple choice question for 1 mark, 2 short answers with reasoning for a total of 4 marks, 3 short answer questions for a total of 9 marks, and 1 long answer question for 6 marks. The chapter covers the following topics:

Chapter 4 Linear Equations in Two Variables
4.1Introduction
4.2Linear Equations
4.3Solution of a Linear Equation
4.4Graph of a Linear Equation in Two Variables
4.5Equations of Lines Parallel to the x-axis and y-axis

In this class 9 chapter 4 maths, there are a total of 4 exercises which consist of 16 questions. NCERT solutions for Class 9 Maths chapter 4 Linear Equations in Two Variables is covering the detailed solutions to each and every question present in the practice exercises. This is an important chapter as this created a foundation for the higher level of algebra. Algebra is a unit in class 9 maths which holds 20 marks in the final examination.

Interested students can practice class 9 maths ch 4 question answer using the links given below.

NCERT Solutions for Class 9 Maths - Chapter Wise

Chapter No.
Chapter Name
Chapter 1
Chapter 2
Chapter 3
Chapter 4
Linear Equations In Two Variables
Chapter 5
Chapter 6
Chapter 7
Chapter 8
Chapter 9
Chapter 10
Chapter 11
Chapter 12
Chapter 13
Chapter 14
Chapter 15

NCERT Solutions for Class 9 - Subject Wise

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables - Key Features

Comprehensive and Step-by-Step: These maths chapter 4 class 9 solutions provide comprehensive coverage of the topic, presented in a step-by-step format for easy understanding.

Practice-Driven: They offer a variety of exercises, including illustrative examples and practical applications, along with additional practice questions to strengthen problem-solving skills.

Accurate and Exam-Oriented: The ch 4 maths class 9 solutions are accurate, exam-oriented, and readily accessible, making them an ideal resource for self-study and preparation for class assessments and board exams.

NCERT Books and NCERT Syllabus

Keep Working Hard & Happy Learning!

Frequently Asked Questions (FAQs)

1. What are the important topics in chapter Linear Equations in Two Variables ?
  • Linear equations, solutions of linear equations
  • The graph of linear equation in two variables,
  • Equations of lines parallel to x-axis and y-axis are the important topics of this chapter.
2. Is it necessary to learn all the questions in linear equations in two variables questions?

Yes, practicing all the questions and formulas related to maths class 9 chapter 4  is essential to perform well in CBSE exams. Careers360 website provides accurate and easy-to-understand solutions, which can be beneficial for students to score higher marks. Apart from exam preparation, these solutions can also assist in solving homework and assignments. Therefore, students can start practicing NCERT Solutions for Class 9 Maths Chapter 4 to improve their overall performance.

3. Where can I find the complete class 9 maths chapter 4 solutions?

Here you will get the detailed NCERT solutions for class 9 maths  by clicking on the link. you can practice linear equations class 9 NCERT solutions which provide you indepth understanding of concepts that ultimately lead to score well in the exam. Also for ease you can study linear equations in two variables class 9 pdf both online and offline mode.

4. How does the NCERT solutions are helpful ?

NCERT solutions are provided in a very detailed manner which will give the conceptual clarity to the students. Also, they can take help from these solutions if they are not able to solve on their own

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20,000 \, \, J - 50,000 \, \, J
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A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

 Option 2)

\; K\;
Option 3)

zero\;

 Option 4)

K/4
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In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

 Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced
Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

 Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .
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How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

 Option 2)

3.125 × 10-2
Option 3)

1.25 × 10-2

 Option 4)

2.5 × 10-2
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If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

 Option 2)

increase two fold
Option 3)

remain unchanged

 Option 4)

be a function of the molecular mass of the substance.
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With increase of temperature, which of these changes?

Option 1)

Molality

 Option 2)

Weight fraction of solute
Option 3)

Fraction of solute present in water

 Option 4)

Mole fraction.
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Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

 Option 2)

6.023 × 1022
Option 3)

half that in 8 g He

 Option 4)

558.5 × 6.023 × 1023
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A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

 Option 2)

more than 3 but less than 6
Option 3)

more than 6 but less than 9

 Option 4)

more than 9
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